Critical Behavior of the Annealed Ising Model on Random Regular Graphs

Abstract

In Giardinà et al. (ALEA Lat Am J Probab Math Stat 13(1):121–161, 2016), the authors have defined an annealed Ising model on random graphs and proved limit theorems for the magnetization of this model on some random graphs including random 2-regular graphs. Then in Can (Annealed limit theorems for the Ising model on random regular graphs, arXiv:1701.08639, 2017), we generalized their results to the class of all random regular graphs. In this paper, we study the critical behavior of this model. In particular, we determine the critical exponents and prove a non standard limit theorem stating that the magnetization scaled by $n^{3/4}$ converges to a specific random variable, with n the number of vertices of random regular graphs.

Appendix: Proof of (54) and (55)

We will repeat some computations in [9] and use Lemma 2.1 to prove these claims.

1.1 Proof of (54)

Using Stirling’s formula, we have

$$\begin{aligned} \left( {\begin{array}{c}n\\ j\end{array}}\right) = \left( \frac{1}{\sqrt{2 \pi }} + o(1) \right) \sqrt{\frac{n}{j(n-j)}} \exp \left( n I \left( \frac{j}{n}\right) \right) , \end{aligned}$$

with

$$\begin{aligned} I(t)=(t-1) \log (1-t) -t \log t. \end{aligned}$$

Therefore using (53), we get

$$\begin{aligned} \frac{x_j(n)}{x_{j_*}(n)}= & {} (1+o(1)) \sqrt{\frac{j_*(n-j_*)}{j(n-j)} } \exp \Bigg ( n \left[ I \left( \frac{j}{n}\right) - I \left( \frac{j_*}{n}\right) \right] \nonumber \\&+\, \log g(dj,dn) - \log g(dj_*,dn) \Bigg ) \nonumber \\= & {} (1+o(1))\sqrt{\frac{j_*(n-j_*)}{j(n-j)} } \exp \left( n \left[ I \left( \frac{j}{n}\right) - dF \left( \frac{j}{n}\right) \right] \right. \nonumber \\&\left. -\, n \left[ I \left( \frac{j_*}{n}\right) - d F \left( \frac{j_*}{n}\right) \right] + \left[ \log g(dj,dn)-ndF \left( \frac{j}{n}\right) \right] \right. \nonumber \\&\left. -\, \left[ \log g(dj_*,dn)-ndF \left( \frac{j_*}{n}\right) \right] \right) \nonumber \\= & {} (1+o(1))\sqrt{\frac{j_*(n-j_*)}{j(n-j)} } \exp \left( n \left[ H \left( \frac{j}{n}\right) - H \left( \frac{j_*}{n}\right) \right] \right. \nonumber \\&\left. + \left[ \log g(dj,dn)-ndF \left( \frac{j}{n}\right) \right] - \left[ \log g(dj_*,dn)-ndF \left( \frac{j_*}{n}\right) \right] \right) , \end{aligned}$$

which yields (54).

1.2 Proof of (55)

Since H(t) attains the maximum at a unique point $\tfrac{1}{2}$, there exists a positive constant $\varepsilon$, such that for all $\delta \le \varepsilon$,

$$\begin{aligned} \max _{|t-\tfrac{1}{2}|\ge \delta } H(t) = \max \{H(\tfrac{1}{2} \pm \delta )\}. \end{aligned}$$

Hence for n large enough (such that $n^{-1/6} \le \varepsilon$), we have for all $|j-j_*| > n^{5/6}$,

$$\begin{aligned} H(j/n)-H(j_*/n) \le \max \{H(j_*/n \pm n^{-1/6})- H(j_*/n) \}. \end{aligned}$$

(68)

Using the same arguments for (58), we can prove that

$$\begin{aligned} H\big (j_*/n \pm n^{-1/6}\big )- H(j_*/n) = \alpha _* n^{-2/3} + o\big (n^{-2/3}\big ). \end{aligned}$$

Therefore

$$\begin{aligned} n \Big (H\big (j_*/n \pm n^{-1/6}\big )- H(j_*/n)\Big ) = \alpha _* n^{1/3} +o\big (n^{1/3}\big ). \end{aligned}$$

(69)

Using $\alpha _*$ and (68), (69), we have for n large enough and $|j-j_*| > n^{5/6}$,

$$\begin{aligned} n \big (H(j/n)-H(j_*/n)\big ) \le \frac{\alpha _* n^{1/3}}{2}. \end{aligned}$$

(70)

On the other hand, for all j

$$\begin{aligned} \sqrt{\frac{j_*(n-j_*)}{j(n-j)}} \le \sqrt{n}. \end{aligned}$$

(71)

It follows from (54), (57), (70) and (71) that for n large enough and $|j-j_*| > n^{5/6}$,

$$\begin{aligned} x_j (n) \le x_{j_*} (n) \sqrt{n} \exp \left( \frac{\alpha _* n^{1/3}}{2} \right) \le x_{j_*} (n) n^{-6}, \end{aligned}$$

since $\alpha _*<0$. Therefore

$$\begin{aligned} \bar{A}_n:=\sum _{|j-j_*| > n^{5/6}} x_j(n) \le x_{j_*}(n) n^{-5}, \end{aligned}$$

(72)

here we recall that $x_j(n)=0$ for all $j<0$ or $j >n$. Similarly, for n large enough and $|j-j_*|> n^{5/6}$,

$$\begin{aligned} \exp \left( \frac{r(2j-n)}{n^{3/4}}\right) x_j(n)\le & {} x_{j_*}(n) \sqrt{n} \exp \left( \frac{|r(2j-n)|}{n^{3/4}}\right) \exp \left( \frac{\alpha _* n^{1/3}}{2}\right) \\\le & {} x_{j_*}(n) \sqrt{n} \exp \left( |r| n^{1/4}+ \frac{\alpha _* n^{1/3}}{2}\right) \\\le & {} x_{j_*}(n) n^{-6}. \end{aligned}$$

Hence

$$\begin{aligned} \bar{B}_n:= \sum _{|j-j_*| > n^{5/6}} \exp \left( \frac{r(2j-n)}{n^{3/4}}\right) x_j(n) \le x_{j_*}(n) n^{-5}. \end{aligned}$$

(73)

Since all the terms $(x_j(n))$ are non negative,

$$\begin{aligned} \hat{A}_n=\sum _{|j-j_*| \le n^{5/6}} x_j (n) \ge x_{j_*} (n), \end{aligned}$$

(74)

and

$$\begin{aligned} \hat{B}_n = \sum _{|j-j_*| \le n^{5/6}} \exp \left( \frac{r(2j-n)}{n^{3/4}}\right) x_j (n)\ge & {} \exp \left( \frac{r(2j_*-n)}{n^{3/4}}\right) x_{j_*} (n) \nonumber \\\ge & {} \exp \left( -\frac{|r|}{n^{3/4}} \right) x_{j_*}(n) \ge \frac{x_{j_*}(n)}{2}, \end{aligned}$$

(75)

for n large enough and r fixed. Finally, combining (72), (73), (74) and (75) yields that

$$\begin{aligned} \frac{A_n}{B_n} = \frac{\hat{A}_n + \bar{A}_n}{\hat{B}_n + \bar{B}_n} = \frac{\hat{A}_n}{\hat{B}_n } + o\big (n^{-2}\big ). \end{aligned}$$